Cambridge A Level Mathematics 9709 — 2011 May/June Paper 6 · Variant 3

2 Fahad has 4 different coloured pairs of shoes (white, red, blue and black), 3 different coloured pairs of jeans (blue, black and brown) and 7 different coloured tee shirts (red, orange, yellow, blue, green, white and purple). (i) Fahad chooses an outfit consisting of one pair of shoes, one pair of jeans and one tee shirt. How many different outfits can he choose? [1] (ii) How many different ways can Fahad arrange his 3 jeans and 7 tee shirts in a row if the two blue items are not next to each other? [2] Fahad also has 9 different books about sport. When he goes on holiday he chooses at least one of these books to take with him. (iii) How many different selections are there if he can take any number of books ranging from just one of them to all of them? [3]

Mark scheme: Σx A 2 2 (ii) − 6.3 = 1.9252 M1 Attempt to find ΣxA using correct 9 variance formula ΣxA 2 = 150 A1 Correct ΣxA 2 1500. + 352 2 − .4017 = 4.780 M1 Using 352 + their 150 in correct variance 24 formula sd = 2.19 A1 [4] Correct answer 2 (i) 4 × 3 × 7 = 84 B1 [1] Correct answer (ii) 10! – 9! × 2 B1 10! − k × 9! seen oe = 2903040 (2900000) B1 [2] Correct answer OR 8! × 9 × 8 B1 8! × 9 × l seen oe = 2903040 (2900000) B1 Correct answer (iii) 9C1 + 9C2 + ... + 9C9 M1 Using combinations M1 Adding 9 combinations = 511 A1 [3] Correct answer OR 29 – 1 M1 29 seen M1 Subtracting 1 = 511 A1 Correct answer 3 (i) medianA < 35 or 20 ≤ medianA < 35 or B1 Correct numerical statement re medianA or medianA = 33.0/33.1/33.5/33.6 medianB or medianB ≥ 50 or 50 ≤ medianB < 70 or B1 [2] Correct numerical statement re other medianB = 51.7/51.9/52.2/52.4 median and a conclusion medianB > medianA OR A has 66 cand 50 < mark < 100, so medA < 50 B1 As before or B has 156 cand 50 < mark < 100, so medB > 50 medianB > medianA B1 As before (ii) 159 – 68 = 91 B1 [1] Correct final answer  5.4 × 25 + 14 5. × 43 + 27 × 91  (iii) mean=   / 300 M1 Using an attempt at mid-points, not end  + ..... + 84 5. × 40  points or class widths M1 Using an attempt at frequencies, not cum freqs M1 Sum of 6 prods, correct freqs, divided by 300 = 11270 / 300 = 37.6 A1 [4] Correct answer GCE AS/A LEVEL – May/June 2011 9709 63 4 (i) (a) P(final score is 12) = P(6, 6) = 1/36 B1 [1] Correct answer (b) P[(1,5) + (1,4) + (2,3) + (3,2) + (4,1)] M1 Considering P(1, 5) M1 Considering P[(1,4) + (2,3) + (3,2) + (4,1)] = 5/36 A1 [3] Correct answer (ii) P(A) = 1/6 P(B) = P[(1,5) + (2,4) + (3,3) + (4, 2) + (5,1)] = 5/36 B1 Any two of P(A), P(B) and P(C) correct P(C) = 1 – P(O, O) = 3/4 B1 Third probability correct P(A and B) = P(1 and 5) = 1/36 ≠ P(A) × P(B) M1 Numerical attempt to compare P(X and Y) P(A and C) = P[(2,5) + (4,5) + (6,5)] = 3/36 with P(X) × P(Y), must be three positive ≠ P(A) × P(C) probs P(B and C) = P[(2,4) + (4,2)] = 2/36 ≠ P(B) × P(C) None are independent. A1√ One correct comparison and conclusion, ft their probabilities A1 [5] Correct conclusion(s) following legitimate working 5 (i) z = ± 1.751 B1 Correct z 20 − µ ± = .1751 M1 Standardising no cc, no sqrt, must be a µ / 4 z-value µ = 13.9 A1 [3] Correct answer 10 − 13.91 (ii) P(X < 10) = P(z < ± ) M1 Standardising attempt with 10, their µ and 13.91 / 4 their µ/4, no cc, no sqrt = P(z < −1.124) M1 “Φ + Φ2 – 1”, ft their mean = 1− 0.8694 = 0.131 P(10 < X < 20) = 0.96 – 0.131 = 0.829 or 0.830 A1 [3] Correct answer (iii) µ = 250 × 0.96 = 240 B1 240 and 9.6 or sq rt 9.6 seen unsimplified σ2 = 250 × 0.96 × 0.04 = 9.6  234 5. − 240  P(≥ 235) = 1 − Φ  ±  M1 Standardising, with or without cc, must  6.9  have sq rt in denom M1 Continuity correction 234.5 or 235.5 only = Φ (1.775) M1 Correct region > 0.5, ft their mean = 0.962 A1 [5] Correct answer GCE AS/A LEVEL – May/June 2011 9709 63 6 (i) (0.75)n < 0.06 M1* Equation or inequality with 0.75n and 0.06 or 0.94 seen n > 9.78 M1dep* Attempt at solving by trial and error (can be implied) or using logarithms correctly n = 10 A1 [3] Correct answer (ii) E(X) = 14 × 0.75 or 10.5 M1 Evaluating binomial probability for an Try P(10) = 14C10(0.75)10(0.25)4 = 0.220 integer value directly above or below their mean P(11) = 14C11(0.75)11(0.25)3 = 0.240 M1 Evaluating the other binomial probability (mode is) 11 A1 [3] Correct answer OR M1 Evaluating binomial P(n) and P(n + 1) M1 Evaluating binomial P(10), P(11) and P(12) A1 Correct answer (iii) P(> 11) M1 A binomial term of the form = 14C12(0.75)12(0.25)2 + 14C13(0.75)13(0.25)1 14Cn pn(1 − p)14 – n seen, n ≠ 0 or 14 + (0.75)14 M1 Summing binomial P(12, 13, 14) or P(11, 12, 13, 14,) = 0.281 A1 Correct answer 0.280 – 0.282 P(3) = 5C3 (0.2811)3(0.7189)2 M1 A binomial term of the form 5C3p3(1 − p)2 seen, any p = 0.115 A1 [5] Correct answer

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